Phân tích đa thức thành nhân tử a)x^4-2x^3+2x-1 b)x^4+2x^3+2x^2+2x+1 30/09/2021 Bởi Abigail Phân tích đa thức thành nhân tử a)x^4-2x^3+2x-1 b)x^4+2x^3+2x^2+2x+1
Đáp án: Giải thích các bước giải: \[\begin{array}{l} a)\,\,\,{x^4} – 2{x^3} + 2x – 1\\ = \left( {{x^4} – 1} \right) – 2x\left( {{x^2} – 1} \right)\\ = \left( {{x^2} – 1} \right)\left( {{x^2} + 1} \right) – 2x\left( {{x^2} – 1} \right)\\ = \left( {{x^2} – 1} \right)\left( {{x^2} – 2x + 1} \right)\\ = \left( {x – 1} \right)\left( {x + 1} \right){\left( {x – 1} \right)^2}\\ = \left( {x + 1} \right){\left( {x – 1} \right)^3}\\ b)\,\,\,{x^4} + 2{x^3} + 2{x^2} + 2x + 1\\ = \left( {{x^4} + 2{x^2} + 1} \right) + \left( {2{x^3} + 2x} \right)\\ = {\left( {{x^2} + 1} \right)^2} + 2x\left( {{x^2} + 1} \right)\\ = \left( {{x^2} + 1} \right)\left( {{x^2} + 1 + 2x} \right)\\ = \left( {{x^2} + 1} \right){\left( {x + 1} \right)^2} \end{array}\] Bình luận
\[\begin{array}{l} a){x^4} – 2{x^3} + 2x – 1\\ = \left( {{x^2} – 1} \right)\left( {{x^2} + 1} \right) – 2x\left( {{x^2} – 1} \right)\\ = \left( {x – 1} \right)\left( {x + 1} \right)\left( {{x^2} + 1} \right) – 2x\left( {x – 1} \right)\left( {x + 1} \right)\\ = \left( {x – 1} \right)\left( {x + 1} \right)\left( {{x^2} + 1 – 2x} \right)\\ = \left( {x – 1} \right)\left( {x + 1} \right){\left( {x – 1} \right)^2}\\ = {\left( {x – 1} \right)^3}\left( {x + 1} \right)\\ b){x^4} + 2{x^3} + 2{x^2} + 2x + 1\\ = \left( {{x^4} + 2{x^3} + {x^2}} \right) + \left( {{x^2} + 2x + 1} \right)\\ = {x^2}\left( {{x^2} + 2x + 1} \right) + \left( {{x^2} + 2x + 1} \right)\\ = {x^2}{\left( {x + 1} \right)^2} + {\left( {x + 1} \right)^2}\\ = {\left( {x + 1} \right)^2}\left( {{x^2} + 1} \right) \end{array}\] Bình luận
Đáp án:
Giải thích các bước giải:
\[\begin{array}{l}
a)\,\,\,{x^4} – 2{x^3} + 2x – 1\\
= \left( {{x^4} – 1} \right) – 2x\left( {{x^2} – 1} \right)\\
= \left( {{x^2} – 1} \right)\left( {{x^2} + 1} \right) – 2x\left( {{x^2} – 1} \right)\\
= \left( {{x^2} – 1} \right)\left( {{x^2} – 2x + 1} \right)\\
= \left( {x – 1} \right)\left( {x + 1} \right){\left( {x – 1} \right)^2}\\
= \left( {x + 1} \right){\left( {x – 1} \right)^3}\\
b)\,\,\,{x^4} + 2{x^3} + 2{x^2} + 2x + 1\\
= \left( {{x^4} + 2{x^2} + 1} \right) + \left( {2{x^3} + 2x} \right)\\
= {\left( {{x^2} + 1} \right)^2} + 2x\left( {{x^2} + 1} \right)\\
= \left( {{x^2} + 1} \right)\left( {{x^2} + 1 + 2x} \right)\\
= \left( {{x^2} + 1} \right){\left( {x + 1} \right)^2}
\end{array}\]
\[\begin{array}{l}
a){x^4} – 2{x^3} + 2x – 1\\
= \left( {{x^2} – 1} \right)\left( {{x^2} + 1} \right) – 2x\left( {{x^2} – 1} \right)\\
= \left( {x – 1} \right)\left( {x + 1} \right)\left( {{x^2} + 1} \right) – 2x\left( {x – 1} \right)\left( {x + 1} \right)\\
= \left( {x – 1} \right)\left( {x + 1} \right)\left( {{x^2} + 1 – 2x} \right)\\
= \left( {x – 1} \right)\left( {x + 1} \right){\left( {x – 1} \right)^2}\\
= {\left( {x – 1} \right)^3}\left( {x + 1} \right)\\
b){x^4} + 2{x^3} + 2{x^2} + 2x + 1\\
= \left( {{x^4} + 2{x^3} + {x^2}} \right) + \left( {{x^2} + 2x + 1} \right)\\
= {x^2}\left( {{x^2} + 2x + 1} \right) + \left( {{x^2} + 2x + 1} \right)\\
= {x^2}{\left( {x + 1} \right)^2} + {\left( {x + 1} \right)^2}\\
= {\left( {x + 1} \right)^2}\left( {{x^2} + 1} \right)
\end{array}\]